Problem 145 - A Complete Resource Book in Chemistry for JEE Main

Question

White phosphorus on reaction with \(\mathrm{NaOH}\) gives \(\mathrm{PH}_{3}\) as one of the products. This is a (a) Dimerization reaction (b) Disproportional reaction (c) Condensation reaction (d) Precipitation reaction

Step-by-Step Solution

Answer

The reaction is a disproportional reaction.

Step 1: Identify the Reactants and Products

The exercise states that white phosphorus reacts with sodium hydroxide (\(\mathrm{NaOH}\)) to give phosphine (\(\mathrm{PH}_{3}\)) as one of the products. Thus, our main focus should be on this reaction and how phosphorus changes its oxidation state.

Step 2: Analyze the Reaction Type

In the reaction, phosphorus starts in a zero oxidation state as white phosphorus (\(\mathrm{P}_4\)) and is converted to phosphine \((\mathrm{PH}_3)\), where phosphorus is in the -3 oxidation state. The other product formed is phosphite ion \((\mathrm{H}_2\mathrm{PO}_2^- )\), where the phosphorus is in the +1 oxidation state. This means phosphorus is both oxidized and reduced, indicative of a disproportional reaction.

Step 3: Confirm the Oxidation State Changes

In a disproportional reaction, the same element undergoes both oxidation and reduction. Here, phosphorus is reduced from 0 to -3 (forming \(\mathrm{PH}_3\)) and is oxidized from 0 to +1 (forming \(\mathrm{H}_2\mathrm{PO}_2^-\)). Thus, all criteria for a disproportional reaction are met.

Key Concepts

White Phosphorus

White phosphorus, often symbolized as \( \text{P}_4 \), is a fascinating allotrope of the element phosphorus. It is known for its unique tetrahedral structure, where four phosphorus atoms are bonded together tightly to form a molecule. This specific arrangement results in high reactivity. The tetrahedral molecules of white phosphorus are only weakly bound to each other, allowing them to react easily with other substances.

One of the most interesting characteristics of white phosphorus is that it is a highly reactive, volatile substance, often stored in water to prevent it from reacting with oxygen in the air, which could lead to spontaneous ignition. White phosphorus is known for glowing in the dark due to its slow reaction with oxygen, emitting a faint greenish light, a phenomenon known as chemiluminescence. Its extreme reactivity with chemicals like \( \text{NaOH} \) makes it an important subject of study in chemistry, as evidenced in the reaction forming phosphine as a product.

Phosphine Formation

The formation of phosphine, or \( \text{PH}_3 \), involves the interesting interaction of white phosphorus with sodium hydroxide. This reaction showcases the prowess of phosphorus in participating in complex chemical transformations. During this reaction, phosphorus in the white phosphorus molecule shifts its oxidation state from 0 to -3, forming phosphine.

Phosphine is a colorless, flammable gas and bears a strong odor reminiscent of rotting fish. Its formation in a reaction where a single element is simultaneously oxidized and reduced is quite distinctive. As part of the disproportionation reaction, one phosphorus atom in \( \text{P}_4 \) is reduced to form \( \text{PH}_3 \), marking a significant change in its chemistry.

• Phosphine serves multiple roles, including an industrial use as a fumigant.
• In chemistry labs, it is also considered a precursor for many phosphorus-containing compounds.
• This ability to form phosphine highlights the versatility of phosphorus

in reacting with various substances to yield different products.

Oxidation States in Chemistry

Understanding oxidation states is crucial in explaining the behavior of elements during chemical reactions. An oxidation state, or number, represents the degree of oxidation or reduction of an element within a compound. It provides insight into the electron transfer processes taking place.

In the case of the white phosphorus and sodium hydroxide reaction, phosphorus starts at an oxidation state of 0 in \( \text{P}_4 \) and undergoes a complex transformation:

• In forming \( \text{PH}_3 \), phosphorus' oxidation state is reduced to -3.
• In forming \( \text{H}_2\text{PO}_2^- \), another phosphorus atom's oxidation state is increased to +1.

This simultaneous increase and decrease in oxidation states are indicative of a disproportional reaction, where the same element undergoes both processes. Understanding these states allows chemists to predict and explain the results of reactions more effectively. Whenever a reaction involves these shifts, it signals a more complex interplay that goes beyond straightforward oxidation or reduction.